Math Problems Vault

Problem 1.124

From the harmonic series

\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}+\cdots

remove all terms that contain the digit 9 in the decimal representation of the denominator. The resulting subseries is convergent.

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Remark.

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A. J. Kempner: Amer. Math. Monthly Vol. 21, pp. 48-50.

Problem 1.125

If all the subseries of a series converge then the series is absolutely convergent.

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Problem 1.126

Let $k$ and $l$ denote positive integers. Must the convergent series $a_1+a_2+a_3+\cdots$ be absolutely convergent if all its subseries of the form

a_k+a_{k+l}+a_{k+2l}+a_{k+3l}+\cdots

(subscripts in arithmetic progression) converge?

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Remark.

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K. Knopp: J. reine angew. Math. Vol. 142, pp. 292-293 (1913).

Problem 1.127

Let $k$ and $l$ be integers, $k \geq 1$ , $l \geq 2$ . Must the convergent series $a_1+a_2+a_3+\cdots$ be absolutely convergent if all its subseries of the form

a_k+a_{kl}+a_{kl^2}+a_{kl^3}+\cdots

(subscripts in geometric progression) are convergent?

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Problem 1.128

Let $\varphi (x)$ denote a polynomial assuming integral values for integral $x$ , $\varphi (x) = c_0 x^l + c_1 x^{l-1} + \cdots$ Assume that the degree is $l \geq 1$ and that the coefficient $c_0$ of $x^l$ is positive ( $c_0>0$ ). The values $\varphi (0), \varphi (1), \varphi (2), \ldots$ form a generalized arithmetic progression of order $l$ ; since $c_0>0$ only a finite number of terms of the progression can be negative. Must a convergent series $a_1+a_2+a_3+\cdots$ converge absolutely if all its subseries whose subscripts form a generalized arithmetic progression

a_{\varphi (0)}+a_{\varphi (1)}+a_{\varphi (2)}+a_{\varphi (3)}+\cdots

(omitting the terms with negative subscripts) converge?

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Remark.

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A polynomial $\varphi (x)$ assuming integral values for every integer $x$ need not itself have integral coefficients -- e.g. $\binom {x}{2} = \frac{x(x-1)}{2}$ is such a polynomial. This is the class of polynomials meant here (cf.Part VIII, Chapter 2).

Problem 1.129

If the series $a_1+a_2+a_3+\cdots$ converges absolutely and if every subseries

a_l+a_{2l}+a_{3l}+\cdots , \qquad l=1,2,3,\ldots

has the sum $0$ then $a_1=a_2=\cdots =0$ .

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Remark.

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A. Haar.

Problem 1.130

Consider the set of points determined by all the subseries of

\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\cdots +\frac{2}{3^n}+\cdots .

This set is perfect and nowhere dense (closed and dense in itself, but nowhere dense in the set of all real numbers).

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Remark.

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We have to consider all the subseries, finite and infinite, including the ``empty'' subseries to which we attribute the sum 0. G. Cantor, cf.E. Hewitt and K. Stromberg: Real and Abstract Analysis. Springer: New York 1965, pp. 70-71.

Problem 1.131

Let the terms of the convergent series

p_1+p_2+p_3+\cdots +p_n+\cdots = s

satisfy the inequalities

p_1 \geq p_2 \geq p_3 \geq \cdots , \qquad 0<p_n \leq p_{n+1}+p_{n+2}+p_{n+3}+\cdots , \quad n=1,2,3,\ldots

Then it is possible to represent any number $\sigma$ in the half-closed interval $0<\sigma \leq s$ by an infinite subseries:

p_{i_1}+p_{i_2}+p_{i_3}+\cdots = \sigma .

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Remark.

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Cf.S. Kakeya: Proc. Tokyo math.-phys. Soc. Ser. 2, Vol. 7, p. 250 (1914); Tôhoku Sc. Rep. Vol. 3, p. 159 (1915).

Problem 1.132

Find the series $p_1+p_2+p_3+\cdots$ that satisfies the conditions

p_1 = \frac{1}{2}, \qquad p_n = p_{n+1}+p_{n+2}+p_{n+3}+\cdots , \qquad n=1,2,3,\ldots ,

and verify that in this case every $\sigma$ mentioned in problem 131 can be represented by one infinite subseries only.

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Problem 1.132.1

By rearranging the factors of the infinite product

\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1+\frac{1}{6}\right) \cdots = P_{1,1}

we obtain the infinite product

P_{p,q} = \left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right) \cdots \left(1+\frac{1}{2p}\right) \times \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \cdots \left(1-\frac{1}{2q+1}\right)\left(1+\frac{1}{2p+2}\right) \cdots

in which blocks of $p$ factors greater than 1 alternate with blocks of $q$ factors smaller than 1. (Factors of the same kind remain in the ``natural'' order.) Show that

P_{p,q} = \sqrt{\frac{p}{q}} \, . \qquad \text{[See Problem 202 of Part II.]}

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Remark.

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G. Pólya, Problem: Amer. Math. Monthly Vol. 51, pp. 533-534 (1944). Solution Amer. Math. Monthly Vol. 53, pp. 279-282 (1946).

Problem 1.132.2

By rearranging the terms of the infinite series

\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots = S_{1,1} = 1-\log 2

we obtain the infinite series

S_{p,q} = \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{2p}-\frac{1}{3}-\frac{1}{5}-\cdots -\frac{1}{2q+1}+\frac{1}{2p+2}+\cdots

in which blocks of $p$ positive terms alternate with blocks of $q$ negative terms. (Terms of the same kind remain in the ``natural'' order, steadily decreasing in absolute value.) Show that

S_{p,q} - S_{1,1} = \frac{1}{2} \log \frac{p}{q} .

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Remark.

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See problem 132.1.